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This is the homework problem: convert $x[n]=je^{j\pi n/8}-je^{-j\pi n/8}$ to a real valued sinusoid.

I understand that $\sin\theta=\dfrac{e^{j\theta}-e^{-j\theta}}{2j}$

In the solution, the answers claim that $x[n]=\dfrac{-e^{j\pi n/8}+e^{-j\pi n/8}}{j}$, and I don't understand how to get from the original

$$x[n]=je^{j\pi n/8}-je^{-j\pi n/8}$$

and arrive at

$$x[n]=\dfrac{-e^{j\pi n/8}+e^{-j\pi n/8}}{j}$$

after which it is easy to see $x[n]=-2\sin(\pi n/8)$

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HINT:

It is based on the fact that

$$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$

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    $\begingroup$ Great hint. Multiply the top and bottom by $j$ $\endgroup$ – Idr Dec 29 '20 at 14:53
  • $\begingroup$ Exactly! Well done. :) $\endgroup$ – Gilles Dec 29 '20 at 16:53
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It's because

$$ j \cdot ( a + b) = -\frac{a + b}{j} $$ which stems from the fact that the imaginary unit $j$ has the property :

$$ j = \frac{-1}{j} $$

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